Solve the matrix equations:
Question: Solve the matrix equations: (i) $\left[\begin{array}{ll}x 1\end{array}\right]\left[\begin{array}{rr}1 0 \\ -2 -3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0$ (ii) $\left[\begin{array}{lll}1 2 1\end{array}\right]\left[\begin{array}{lll}1 2 0 \\ 2 0 1 \\ 1 0 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$ (iii) $[x-5-1]\left[\begin{array}{lll}1 0 2 \\ 0 2 1 \\ 2 0 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$ (iv...
Read More →The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36,
Question: The sum of three numbersa,b,cin A.P. is 18. Ifaandbare each increased by 4 andcis increased by 36, the new numbers form a G.P. Finda,b,c. Solution: Let the first term of the A.P. beaand the common difference bed. a = a , b = a + dandc = a + 2d $a+b+c=18$ $\Rightarrow a+(a+d)+(a+2 d)=18$ $\Rightarrow 3 a+3 d=18$ $\Rightarrow a+d=6$ ....(i) Now, according to the question, a $+4$, a $+d+4$ and $a+2 d+36$ are in G. P. $\therefore(\mathrm{a}+\mathrm{d}+4)^{2}=(\mathrm{a}+4)(\mathrm{a}+2 \ma...
Read More →Find the value of a when the distance between the points (3, a)
Question: Find the value of $a$ when the distance between the points $(3, a)$ and $(4,1)$ is $\sqrt{10}$. Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The distance between two points $(3, a)$ and $(4,1)$ is given as $\sqrt{10}$. Substituting these values in the formula for distance between two points we have, $\sqrt{10}=\sqrt{(3-4)^{2}+(a-1)^{2...
Read More →In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively.
Question: In a circle of radius 5 cm,ABandCDare two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the distance between the chords if they are(i) on the same side of the centre(ii) on the opposite sides of the centre. Solution: We have:(i)LetABandCDbe two chords of a circle such thatABis parallel toCDon the same side of the circle.Given:AB= 8 cm,CD= 6 cm andOB = OD= 5 cmJoinOLandOM. The perpendicular from the centre of a circle to a chord bisects the chord. $\therefore L B=\fra...
Read More →The sum of three numbers which are consecutive terms of an A.P. is 21.
Question: The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers. Solution: Let the first term of an A.P isaand its common difference bed. $\therefore a_{1}+a_{2}+a_{3}=21$ $\Rightarrow a+(a+d)+(a+2 d)=21$ $\Rightarrow 3 a+3 d=21$ $\Rightarrow a+d=7$ ...(i) Now, according to the question: a , $a+d-1$ and $a+2 d+1$ are in $G . P .$ $\Rightarrow(\mathrm...
Read More →Find the distance between the following pair of points:
Question: Find the distance between the following pair of points: (a) (6, 7) and (1, 5)(b) (a+b,b+c) and (ab,cb)(c) (asin, bcos) and (acos ,bsin )(d) (a, 0) and (0,b) Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ (i) The two given points are (6, 7) and (1, 5) The distance between these two points is $d=\sqrt{(-6+1)^{2}+(7+5)^{2}}$ $=\sqrt{(-5)^{...
Read More →Find the value of x for which the matrix product
Question: Find the value ofxfor which the matrix product$\left[\begin{array}{rrr}2 0 7 \\ 0 1 0 \\ 1 -2 1\end{array}\right]\left[\begin{array}{rrr}-x 14 x 7 x \\ 0 1 0 \\ x -4 x -2 x\end{array}\right]$ equal an identity matrix. Solution: Here, $\left[\begin{array}{ccc}2 0 7 \\ 0 1 0 \\ 1 -2 1\end{array}\right]\left[\begin{array}{ccc}-x 14 x 7 x \\ 0 1 0 \\ x -4 x -2 x\end{array}\right]=\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right]$ $\Rightarrow\left[\begin{array}{ccc}-2 x+0+7...
Read More →Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively,
Question: Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers. Solution: Let the first term of an A.P. beaand its common difference bed. $a_{1}+a_{2}+a_{3}=15$ $\Rightarrow a+(a+d)+(a+2 d)=15$ $\Rightarrow 3 a+3 d=15$ $\Rightarrow a+d=5$ ....(i) Now, according to the question: $a+1, a+d+3$ and $a+2 d+9$ are in G.P. $\Rightarrow(\mathrm{a}+\mathrm{d}+3)^{2}=(\mathrm{a}+1)(\mathrm{a}+2 \mathrm{~d}+9)$ $\Rightarrow(5-\mathrm{d}+...
Read More →A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle.
Question: A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle. Solution: LetABbe the chord of the given circle with centreO. The perpendicular distance from the centre of the circle to the chord is 8 cm.JoinOB.ThenOM= 8 cm andAB =30 cm We know that the perpendicular from the centre of a circle to a chord bisects the chord. $\therefore M B=\left(\frac{A B}{2}\right)=\left(\frac{30}{2}\right) \mathrm{cm}=15 \mathrm{~cm}$ From the ri...
Read More →Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.
Question: Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm. Solution: LetABbe the chord of the given circle with centreOand a radius of 5 cm.FromO, drawOMperpendicular toAB.ThenOM= 3 cm andOB= 5 cm From the right ΔOMB, we have:OB2= OM2+ MB2 (Pythagoras theorem)⇒ 52= 32+MB2⇒ 25 = 9 +MB2⇒MB2= (259) = 16 $\Rightarrow M B=\sqrt{16} \mathrm{~cm}=4 \mathrm{~cm}$ Since the perpendicular from the centre of a circle to a chord bisects the chord, we have...
Read More →Find k such that k + 9, k − 6 and 4
Question: Findksuch thatk+ 9,k 6 and 4 form three consecutive terms of a G.P. Solution: k, k+ 9,k6 are in G.P. $\therefore(k-6)^{2}=4(k+9)$ $\Rightarrow k^{2}+36-12 k=4 k+36$ $\Rightarrow k^{2}-16 k=0$ $\Rightarrow k(k-16)=0$ $\Rightarrow k=0,16$ But,k= 0 is not possible. k= 16...
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}2 3 \\ 1 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$, then find $\lambda, \mu$ so that $A^{2}=\lambda A+\mu l$ Solution: Given : $A=\left[\begin{array}{ll}2 3 \\ 1 2\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}2 3 \\ 1 2\end{array}\right]\left[\begin{array}{ll}2 3 \\ 1 2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}4+3 6+6 \\ 2+2 3+4\end{array}\right]$ $\Rightarrow A^{2}=...
Read More →If a, b, c are in G.P., prove that
Question: Ifa,b,care in G.P., prove that $\frac{1}{\log _{a} m}, \frac{1}{\log _{b} m}, \frac{1}{\log _{c} m}$ are in A.P. Solution: a,b,care in G.P. $\therefore b^{2}=a c$ Now taking $\log _{m}$ on both the sides: $\Rightarrow \log _{m}(b)^{2}=\log _{m}(a c)$ $\Rightarrow 2 \log _{m}(b)=\log _{m} a+\log _{m}(c)$ $\Rightarrow \frac{2}{\log _{b}(m)}=\frac{1}{\log _{a}(m)}+\frac{1}{\log _{c}(m)}$ Thus, $\frac{1}{\log _{\mathrm{a}}(\mathrm{m})}, \frac{1}{\log _{\mathrm{b}}(\mathrm{m})}$ and $\frac{...
Read More →A chord of length 16 cm is drawn in a circle of radius 10 cm.
Question: A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the circle. Solution: LetABbe the chord of the given circle with centreOand a radius of 10 cm.ThenAB=16 cm andOB= 10 cm FromO, drawOMperpendicular toAB.We know that the perpendicular from the centre of a circle to a chord bisects the chord. $\therefore B M=\left(\frac{16}{2}\right) \mathrm{cm}=8 \mathrm{~cm}$ In the right ΔOMB, we have:OB2= OM2+ MB2(Pythagoras theorem)⇒ 102=O...
Read More →If a, b, c are in G.P., prove that log a, log b,
Question: Ifa,b,care in G.P., prove that loga, logb, logcare in A.P. Solution: a ,b and c are in G.P. $\therefore b^{2}=a c$ Now, taking $\log$ on both the sides: $\Rightarrow \log (b)^{2}=\log a c$ $\Rightarrow 2 \log b=\log a+\log c$ Thus, $\log a, \log b$ and $\log c$ are in A.P....
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}1 2 \\ 2 1\end{array}\right], f(x)=x^{2}-2 x-3$, show that $f(A)=0$ Solution: Here, $f(x)=x^{2}-2 x-3$ $\Rightarrow f(A)=A^{2}-2 A-3 I_{2}$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1 2 \\ 2 1\end{array}\right]\left[\begin{array}{ll}1 2 \\ 2 1\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1+4 2+2 \\ 2+2 4+1\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}5 4 \\ 4 5\end{array}\right]$ $f(A)=A^{2}-2 A-3 I_{2}$ $\Ri...
Read More →If S denotes the sum of an infinite G.P. S1 denotes the sum of the squares of its terms,
Question: If $S$ denotes the sum of an infinite G.P. $S_{1}$ denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively $\frac{2 S S_{1}}{S^{2}+S_{1}}$ and $\frac{S^{2}-S_{1}}{S^{2}+S_{1}}$. Solution: $\mathrm{S}=\frac{a}{(1-r)}$ ...(i) And, $\mathrm{S}_{1}=\frac{a^{2}}{\left(1-r^{2}\right)}$ $\Rightarrow \mathrm{S}_{1}=\frac{a^{2}}{(1-r)(1+r)} \quad \cdots \cdots$ (ii) Now, putting the value of $a$ in equation (ii) from equation (i): $\mathrm{S...
Read More →Solve this
Question: If $A=\left[\begin{array}{rr}1 0 \\ -1 7\end{array}\right]$, find $k$ such that $A^{2}-8 A+k i=0$ Solution: Given: $A=\left[\begin{array}{cc}1 0 \\ -1 7\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{cc}1 0 \\ -1 7\end{array}\right]\left[\begin{array}{cc}1 0 \\ -1 7\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}1-0 0+0 \\ -1-7 0+49\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{cc}1 0 \\ -8 49\end{array}\right]$ $A^{2}-8 A+k I=0$ $\...
Read More →Show that in an infinite G.P. with common ratio r (|r| < 1),
Question: Show that in an infinite G.P. with common ratior(|r| 1), each term bears a constant ratio to the sum of all terms that follow it. Solution: Let us take a G.P. with terms $a_{1}, a_{2}, a_{3}, a_{4}, \ldots \infty$ and common ratio $r(|r|1)$. Also, let us take the sum of all the terms following each term to be $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}, \mathrm{~S}_{4}, \ldots$ Now, $\mathrm{S}_{1}=\frac{a_{2}}{(1-r)}=\frac{a r}{(1-r)}$ $S_{2}=\frac{a_{3}}{(1-r)}=\frac{a r^{2}}{(...
Read More →If a two digit number is chosen at random, then the
Question: If a two digit number is chosen at random, then the probability that the number chosen is a multiple of 3, is (a) $\frac{3}{10}$ (b) $\frac{29}{100}$ (c) $\frac{1}{3}$ (d) $\frac{7}{25}$ Solution: GIVEN: A two digit number is chosen at random TO FIND: Probability that the number chosen is a multiple of 3 Total two digit numbers is 90 Two digit Numbers multiple of 3 are 12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99 Hence total two digit number...
Read More →From the letters of the word ''MOBILE",
Question: From the letters of the word ''MOBILE", a letter is selected. The probability that the letter is a vowel, is (a) $\frac{1}{3}$ (b) $\frac{3}{7}$ (c) $\frac{1}{6}$ (d) $\frac{1}{2}$ Solution: GIVEN: A letter is selected from the word MOBILE TO FIND: Probability that the letter chosen is a vowel Total letter in the word MOBILE is 6 Vowels in the word MOBILE are O.I,E Hence total number of favorable outcome is 3 i.e. O.I,E We know that PROBABILITY = Hence probability that the letter chose...
Read More →A month is selected at random in a year.
Question: A month is selected at random in a year. The probability that it is March or October, is (a) $\frac{1}{12}$ (b) $\frac{1}{6}$ (c) $\frac{3}{4}$ (d) None of these Solution: GIVEN: A month is selected at random in a year. TO FIND: Probability that it is March or October Total months in a year is 12 Hence total number of favorable outcome is 2 i.e. March or October We know that PROBABILITY = Hence probability that the month selected is March or October is equal to Hence the correct option...
Read More →The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of the succeeding terms.
Question: The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of the succeeding terms. Find the G.P. Solution: Let the first term beaand the common difference ber. $\therefore a_{1}+a_{2}=5$ $\Rightarrow a+a r=5$ ....(i) Also, $a_{n}=3\left[a_{n+1}+a_{n+2}+a_{n+3}+\ldots \infty\right] \forall n \in N$ $\Rightarrow a r^{n-1}=3\left[a r^{n+1}+a r^{n+2}+a r^{n+3}+\ldots \infty\right]$ $\Rightarrow a r^{n-1}=\frac{3 a r^{n}}{1-r}$ $\Rightarrow 1-r=3 r$ $\Rightarr...
Read More →if
Question: If $A=\left[\begin{array}{rr}3 -2 \\ 4 -2\end{array}\right]$, find $k$ such that $A^{2}=k A-2 / 2$ Solution: Given: $A=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$ Now, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}9-8 -6+4 \\ 12-8 -8+4\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ll}1 -2 \\ 4 4\end{array}\right]$ $A^{2}=k A-2 I_{...
Read More →What is the probability that a leap year has 52 Mondays?
Question: What is the probability that a leap year has 52 Mondays? (a) $\frac{2}{7}$ (b) $\frac{4}{7}$ (c) $\frac{5}{7}$ (d) $\frac{6}{7}$ Solution: GIVEN: A leap year TO FIND: Probability that a leap year has 52 Mondays. Total number of days in leap year is 366days Hence number of weeks in a leap year is In a leap year we have 52 complete weeks and 2 day which can be any pair of the day of the week i.e. (Sunday, Monday) (Monday, Tuesday) (Tuesday, Wednesday) (Wednesday, Thursday) (Thursday, Fri...
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